Longest subarray with GCD greater than 1

Given an array arr[] consisting of N integers, the task is to find the maximum length of subarray having the Greatest Common Divisor (GCD) of all the elements greater than 1.

Examples:

Input: arr[] = {4, 3, 2, 2}
Output: 2
Explanation:
Consider the subarray {2, 2} having GCD as 2(> 1) which is of maximum length.

Input: arr[] = {410, 52, 51, 180, 222, 33, 33}
Output: 5

Naive Approach: The given problem can be solved by generating all possible subarrays of the given array and keep track of the maximum length of the subarray with GCD greater than 1. After checking for all the subarrays, print the maximum length obtained.

Below is the implementation of the above approach:

C++

// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find maximum length of
// the subarray having GCD > one
void maxSubarrayLen(int arr[], int n)
{
    // Stores maximum length of subarray
    int maxLen = 0;

    // Loop to iterate over all subarrays
    // starting from index i
    for (int i = 0; i < n; i++) {

        // Find the GCD of subarray
        // from i to j
        int gcd = 0;
        for (int j = i; j < n; j++) {

            // Calculate GCD
            gcd = __gcd(gcd, arr[j]);

            // Update maximum length
            // of the subarray
            if (gcd > 1)
                maxLen = max(maxLen, j - i + 1);
            else
                break;
        }
    }

    // Print maximum length
    cout << maxLen;
}

// Driver Code
int main()
{
    int arr[] = { 410, 52, 51, 180,
                  222, 33, 33 };
    int N = sizeof(arr) / sizeof(int);
    maxSubarrayLen(arr, N);

    return 0;
}

Java

// Java code for above approach
import java.util.*;

class GFG{

    // Recursive function to return gcd of a and b
    static int __gcd(int a, int b)
    {
        // Everything divides 0
        if (a == 0)
          return b;
        if (b == 0)
          return a;

        // base case
        if (a == b)
            return a;

        // a is greater
        if (a > b)
            return __gcd(a-b, b);
        return __gcd(a, b-a);
    }

// Function to find maximum length of
// the subarray having GCD > one
static void maxSubarrayLen(int arr[], int n)
{
    // Stores maximum length of subarray
    int maxLen = 0;

    // Loop to iterate over all subarrays
    // starting from index i
    for (int i = 0; i < n; i++) {

        // Find the GCD of subarray
        // from i to j
        int gcd = 0;
        for (int j = i; j < n; j++) {

            // Calculate GCD
            gcd = __gcd(gcd, arr[j]);

            // Update maximum length
            // of the subarray
            if (gcd > 1)
                maxLen = Math.max(maxLen, j - i + 1);
            else
                break;
        }
    }

    // Print maximum length
    System.out.print(maxLen);
}


// Driver Code
public static void main(String[] args)

{
    int arr[] = { 410, 52, 51, 180,
                  222, 33, 33 };
    int N = arr.length;
    maxSubarrayLen(arr, N);
}
}

// This code is contributed by avijitmondal1998.

Python3

# Python program of the above approach
def __gcd(a, b):
  if (b == 0): return a;
  return __gcd(b, a % b);


# Function to find maximum length of
# the subarray having GCD > one
def maxSubarrayLen(arr, n) :

  # Stores maximum length of subarray
  maxLen = 0;

  # Loop to iterate over all subarrays
  # starting from index i
  for i in range(n):

    # Find the GCD of subarray
    # from i to j
    gcd = 0;
    for j in range(i, n):

      # Calculate GCD
      gcd = __gcd(gcd, arr[j]);

      # Update maximum length
      # of the subarray
      if (gcd > 1):
           maxLen = max(maxLen, j - i + 1);
      else: break;

  # Print maximum length
  print(maxLen);

# Driver Code
arr = [410, 52, 51, 180, 222, 33, 33];
N = len(arr)
maxSubarrayLen(arr, N);

# This code is contributed by gfgking.

C#

// C# code for the above approach
using System;
using System.Collections.Generic;
public class GFG
{

  // Function to find maximum length of
  // the subarray having GCD > one
  static int __gcd(int a, int b)
  {

    // Everything divides 0
    if (a == 0)
      return b;
    if (b == 0)
      return a;

    // base case
    if (a == b)
      return a;

    // a is greater
    if (a > b)
      return __gcd(a - b, b);

    return __gcd(a, b - a);
  }
  static void maxSubarrayLen(int []arr, int n)
  {

    // Stores maximum length of subarray
    int maxLen = 0;

    // Loop to iterate over all subarrays
    // starting from index i
    for (int i = 0; i < n; i++) {

      // Find the GCD of subarray
      // from i to j
      int gcd = 0;
      for (int j = i; j < n; j++) {

        // Calculate GCD
        gcd = __gcd(gcd, arr[j]);

        // Update maximum length
        // of the subarray
        if (gcd > 1)
          maxLen = Math.Max(maxLen, j - i + 1);
        else
          break;
      }
    }

    // Print maximum length
    Console.Write(maxLen);
  }

  // Driver Code
  static public void Main (){
    int []arr = { 410, 52, 51, 180,
                 222, 33, 33 };
    int N = arr.Length;
    maxSubarrayLen(arr, N);
  }
}

// This code is contributed by Potta Lokesh

Javascript

<script>
// Javascript program of the above approach
function __gcd(a, b) {
  if (b == 0) return a;
  return __gcd(b, a % b);
}

// Function to find maximum length of
// the subarray having GCD > one
function maxSubarrayLen(arr, n)
{

  // Stores maximum length of subarray
  let maxLen = 0;

  // Loop to iterate over all subarrays
  // starting from index i
  for (let i = 0; i < n; i++)
  {

    // Find the GCD of subarray
    // from i to j
    let gcd = 0;
    for (let j = i; j < n; j++)
    {

      // Calculate GCD
      gcd = __gcd(gcd, arr[j]);

      // Update maximum length
      // of the subarray
      if (gcd > 1) maxLen = Math.max(maxLen, j - i + 1);
      else break;
    }
  }

  // Print maximum length
  document.write(maxLen);
}

// Driver Code
let arr = [410, 52, 51, 180, 222, 33, 33];
let N = arr.length;
maxSubarrayLen(arr, N);

// This code is contributed by _saurabh_jaiswal.
</script>

Output:

5

Time Complexity: O(N²)
Auxiliary Space: O(1)

Efficient Approach: The above can also be optimized using the Sliding Window Technique and Segment Tree. Suppose L denotes the first index, R denotes the last index and X denotes the size of the current window, then there are two possible cases as discussed below:

• The GCD of the current window is 1. In this case, move to the next window of size X (i.e. L + 1 to R + 1).
• The GCD of the current window is greater than 1. In this case, increase the size of the current window by one, (i.e., L to R + 1).

Therefore, to implement the above idea, a Segment Tree can be used to find the GCD of the given index ranges of the array efficiently.

Below is the implementation of the above approach:

C++

// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to build the Segment Tree
// from the given array to process
// range queries in log(N) time
void build_tree(int* b, vector<int>& seg_tree,
                int l, int r, int vertex)
{
    // Termination Condition
    if (l == r) {
        seg_tree[vertex] = b[l];
        return;
    }

    // Find the mid value
    int mid = (l + r) / 2;

    // Left and Right  Recursive Call
    build_tree(b, seg_tree, l, mid,
               2 * vertex);
    build_tree(b, seg_tree, mid + 1,
               r, 2 * vertex + 1);

    // Update the Segment Tree Node
    seg_tree[vertex] = __gcd(seg_tree[2 * vertex],
                             seg_tree[2 * vertex + 1]);
}

// Function to return the GCD of the
// elements of the Array from index
// l to index r
int range_gcd(vector<int>& seg_tree, int v,
              int tl, int tr,
              int l, int r)
{
    // Base Case
    if (l > r)
        return 0;
    if (l == tl && r == tr)
        return seg_tree[v];

    // Find the middle range
    int tm = (tl + tr) / 2;

    // Find the GCD and return
    return __gcd(
        range_gcd(seg_tree, 2 * v, tl,
                  tm, l, min(tm, r)),

        range_gcd(seg_tree, 2 * v + 1,
                  tm + 1, tr,
                  max(tm + 1, l), r));
}

// Function to print maximum length of
// the subarray having GCD > one
void maxSubarrayLen(int arr[], int n)
{
    // Stores the Segment Tree
    vector<int> seg_tree(4 * (n) + 1, 0);

    // Function call to build the
    // Segment tree from array arr[]
    build_tree(arr, seg_tree, 0, n - 1, 1);

    // Store maximum length of subarray
    int maxLen = 0;

    // Starting and ending pointer of
    // the current window
    int l = 0, r = 0;

    while (r < n && l < n) {

        // Case where the GCD of the
        // current window is 1
        if (range_gcd(seg_tree, 1, 0,
                      n - 1, l, r)
            == 1) {
            l++;
        }

        // Update the maximum length
        maxLen = max(maxLen, r - l + 1);
        r++;
    }

    // Print answer
    cout << maxLen;
}

// Driver Code
int main()
{
    int arr[] = { 410, 52, 51, 180, 222, 33, 33 };
    int N = sizeof(arr) / sizeof(int);
    maxSubarrayLen(arr, N);

    return 0;
}

Java

// Java program of the above approach
class GFG{

// Function to build the Segment Tree
// from the given array to process
// range queries in log(N) time
static void build_tree(int[] b, int [] seg_tree,
                int l, int r, int vertex)
{

    // Termination Condition
    if (l == r) {
        seg_tree[vertex] = b[l];
        return;
    }

    // Find the mid value
    int mid = (l + r) / 2;

    // Left and Right  Recursive Call
    build_tree(b, seg_tree, l, mid,
               2 * vertex);
    build_tree(b, seg_tree, mid + 1,
               r, 2 * vertex + 1);

    // Update the Segment Tree Node
    seg_tree[vertex] = __gcd(seg_tree[2 * vertex],
                             seg_tree[2 * vertex + 1]);
}
static int __gcd(int a, int b)
{
    return b == 0? a:__gcd(b, a % b);
}

// Function to return the GCD of the
// elements of the Array from index
// l to index r
static int range_gcd(int [] seg_tree, int v,
              int tl, int tr,
              int l, int r)
{
    // Base Case
    if (l > r)
        return 0;
    if (l == tl && r == tr)
        return seg_tree[v];

    // Find the middle range
    int tm = (tl + tr) / 2;

    // Find the GCD and return
    return __gcd(
        range_gcd(seg_tree, 2 * v, tl,
                  tm, l, Math.min(tm, r)),

        range_gcd(seg_tree, 2 * v + 1,
                  tm + 1, tr,
                  Math.max(tm + 1, l), r));
}

// Function to print maximum length of
// the subarray having GCD > one
static void maxSubarrayLen(int arr[], int n)
{
    // Stores the Segment Tree
    int []seg_tree = new int[4 * (n) + 1];

    // Function call to build the
    // Segment tree from array arr[]
    build_tree(arr, seg_tree, 0, n - 1, 1);

    // Store maximum length of subarray
    int maxLen = 0;

    // Starting and ending pointer of
    // the current window
    int l = 0, r = 0;

    while (r < n && l < n) {

        // Case where the GCD of the
        // current window is 1
        if (range_gcd(seg_tree, 1, 0,
                      n - 1, l, r)
            == 1) {
            l++;
        }

        // Update the maximum length
        maxLen = Math.max(maxLen, r - l + 1);
        r++;
    }

    // Print answer
    System.out.print(maxLen);
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 410, 52, 51, 180, 222, 33, 33 };
    int N = arr.length;
    maxSubarrayLen(arr, N);

}
}

// This code is contributed by shikhasingrajput

Python3

# Python3 program of the above approach

# Function to build the Segment Tree
# from the given array to process
# range queries in log(N) time
def build_tree(b, seg_tree, l, r, vertex):
    # Termination Condition
    if (l == r):
        seg_tree[vertex] = b[l]
        return

    # Find the mid value
    mid = int((l + r) / 2)

    # Left and Right  Recursive Call
    build_tree(b, seg_tree, l, mid, 2 * vertex)
    build_tree(b, seg_tree, mid + 1, r, 2 * vertex + 1)

    # Update the Segment Tree Node
    seg_tree[vertex] = __gcd(seg_tree[2 * vertex], seg_tree[2 * vertex + 1])

def __gcd(a, b):
    if b == 0:
        return b
    else:
        return __gcd(b, a % b)

# Function to return the GCD of the
# elements of the Array from index
# l to index r
def range_gcd(seg_tree, v, tl, tr, l, r):
    # Base Case
    if (l > r):
        return 0
    if (l == tl and r == tr):
        return seg_tree[v]

    # Find the middle range
    tm = int((tl + tr) / 2)

    # Find the GCD and return
    return __gcd(range_gcd(seg_tree, 2 * v, tl, tm, l, min(tm, r)),
        range_gcd(seg_tree, 2 * v + 1, tm + 1, tr, max(tm + 1, l), r))

# Function to print maximum length of
# the subarray having GCD > one
def maxSubarrayLen(arr, n):
    # Stores the Segment Tree
    seg_tree = [0]*(4*n + 1)

    # Function call to build the
    # Segment tree from array []arr
    build_tree(arr, seg_tree, 0, n - 1, 1)

    # Store maximum length of subarray
    maxLen = 0

    # Starting and ending pointer of
    # the current window
    l, r = 0, 0

    while (r < n and l < n):
        # Case where the GCD of the
        # current window is 1
        if (range_gcd(seg_tree, 1, 0, n - 1, l, r) == 1):
            l+=1

        # Update the maximum length
        maxLen = max(maxLen, r - l - 1)
        r+=1

    # Print answer
    print(maxLen, end = "")

arr = [ 410, 52, 51, 180, 222, 33, 33 ]
N = len(arr)
maxSubarrayLen(arr, N)

# This code is contributed by divyesh072019.

C#

// C# program of the above approach
using System;
public class GFG
{

// Function to build the Segment Tree
// from the given array to process
// range queries in log(N) time
static void build_tree(int[] b, int [] seg_tree,
                int l, int r, int vertex)
{

    // Termination Condition
    if (l == r) {
        seg_tree[vertex] = b[l];
        return;
    }

    // Find the mid value
    int mid = (l + r) / 2;

    // Left and Right  Recursive Call
    build_tree(b, seg_tree, l, mid,
               2 * vertex);
    build_tree(b, seg_tree, mid + 1,
               r, 2 * vertex + 1);

    // Update the Segment Tree Node
    seg_tree[vertex] = __gcd(seg_tree[2 * vertex],
                             seg_tree[2 * vertex + 1]);
}
static int __gcd(int a, int b)
{
    return b == 0? a:__gcd(b, a % b);
}

// Function to return the GCD of the
// elements of the Array from index
// l to index r
static int range_gcd(int [] seg_tree, int v,
              int tl, int tr,
              int l, int r)
{
    // Base Case
    if (l > r)
        return 0;
    if (l == tl && r == tr)
        return seg_tree[v];

    // Find the middle range
    int tm = (tl + tr) / 2;

    // Find the GCD and return
    return __gcd(
        range_gcd(seg_tree, 2 * v, tl,
                  tm, l, Math.Min(tm, r)),

        range_gcd(seg_tree, 2 * v + 1,
                  tm + 1, tr,
                  Math.Max(tm + 1, l), r));
}

// Function to print maximum length of
// the subarray having GCD > one
static void maxSubarrayLen(int []arr, int n)
{
    // Stores the Segment Tree
    int []seg_tree = new int[4 * (n) + 1];

    // Function call to build the
    // Segment tree from array []arr
    build_tree(arr, seg_tree, 0, n - 1, 1);

    // Store maximum length of subarray
    int maxLen = 0;

    // Starting and ending pointer of
    // the current window
    int l = 0, r = 0;

    while (r < n && l < n) {

        // Case where the GCD of the
        // current window is 1
        if (range_gcd(seg_tree, 1, 0,
                      n - 1, l, r)
            == 1) {
            l++;
        }

        // Update the maximum length
        maxLen = Math.Max(maxLen, r - l + 1);
        r++;
    }

    // Print answer
    Console.Write(maxLen);
}

// Driver Code
public static void Main(String[] args)
{
    int []arr = { 410, 52, 51, 180, 222, 33, 33 };
    int N = arr.Length;
    maxSubarrayLen(arr, N);

}
}

// This code is contributed by shikhasingrajput

Javascript

<script>
// Javascript program of the above approach

function __gcd(a, b) {
  if (b == 0) return a;
  return __gcd(b, a % b);
}

// Function to build the Segment Tree
// from the given array to process
// range queries in log(N) time
function build_tree(b, seg_tree, l, r, vertex) {
  // Termination Condition
  if (l == r) {
    seg_tree[vertex] = b[l];
    return;
  }

  // Find the mid value
  let mid = Math.floor((l + r) / 2);

  // Left and Right  Recursive Call
  build_tree(b, seg_tree, l, mid, 2 * vertex);
  build_tree(b, seg_tree, mid + 1, r, 2 * vertex + 1);

  // Update the Segment Tree Node
  seg_tree[vertex] = __gcd(seg_tree[2 * vertex], seg_tree[2 * vertex + 1]);
}

// Function to return the GCD of the
// elements of the Array from index
// l to index r
function range_gcd(seg_tree, v, tl, tr, l, r) {
  // Base Case
  if (l > r) return 0;
  if (l == tl && r == tr) return seg_tree[v];

  // Find the middle range
  let tm = Math.floor((tl + tr) / 2);

  // Find the GCD and return
  return __gcd(
    range_gcd(seg_tree, 2 * v, tl, tm, l, Math.min(tm, r)),

    range_gcd(seg_tree, 2 * v + 1, tm + 1, tr, Math.max(tm + 1, l), r)
  );
}

// Function to print maximum length of
// the subarray having GCD > one
function maxSubarrayLen(arr, n) {
  // Stores the Segment Tree
  let seg_tree = new Array(4 * n + 1).fill(0);

  // Function call to build the
  // Segment tree from array arr[]
  build_tree(arr, seg_tree, 0, n - 1, 1);

  // Store maximum length of subarray
  let maxLen = 0;

  // Starting and ending pointer of
  // the current window
  let l = 0,
    r = 0;

  while (r < n && l < n) {
    // Case where the GCD of the
    // current window is 1
    if (range_gcd(seg_tree, 1, 0, n - 1, l, r) == 1) {
      l++;
    }

    // Update the maximum length
    maxLen = Math.max(maxLen, r - l + 1);
    r++;
  }

  // Print answer
  document.write(maxLen);
}

// Driver Code

let arr = [410, 52, 51, 180, 222, 33, 33];
let N = arr.length;
maxSubarrayLen(arr, N);

// This code is contributed by gfgking.
</script>

Output:

5

Time Complexity: O(N*log N)
Auxiliary Space: O(N)

Related Topic: Subarrays, Subsequences, and Subsets in Array